Let $M(X)$ is the set of all probability measures on $(X,\mathcal{B})$. We know $M(X)$ is convex.
Let $O=\{f_n\}$ be a countable dense subset of $C(X)$ and define a meter on $M(X)$ in the following sense:
\begin{align} d_O: M(X) \times M(X) \to \mathbb{R}\\ d_O(\mu,\nu)=\Sigma_{n=1}^{\infty} \frac{|\int f_n d\mu- \int f_n d\nu | }{2^n\| f_n\|} \end{align} It is easy to check that the above map is a metric on $M(X)$. What I'm looking for is that this meter is independence of choice of basis $\{f_n\}$.
To prove this my idea is that we consider another meter say $d_G$ on $M(X)$ defined by countable dense basis $\{g_n\}$ and show that these two meters are equivalent. Any help?
It is easy to show that $d(\mu_j,\mu) \to 0$ iff $\int f_nd\mu_j \to \int f_md\mu$ as $j \to \infty$ for every $n$. If this holds then the ineqyality $$|\int fd\mu_j-\int f d\mu|$$ $$ \leq |\int fd\mu_j-\int f_n d\mu_j|+|\int f_nd\mu_j-\int f_n d\mu|+|\int f_nd\mu-\int f d\mu|$$ $$ \leq \|f_n-f\|+|\int f_nd\mu_j-\int f_n d\mu|+\|f_n-f\|$$ shows that $\int fd\mu_j \to \int fd\mu$ for every $f \in C(X)$. It follows that convergence of $(\mu_j)$ to $\mu$ in one of these metrics implies convergence in any other. Hence the topologies induced by these metrics are all the same.